A360461 T(n,k) is the sum of all the k-th smallest divisors of all positive integers <= n. Irregular triangle read by rows (n>=1, k>=1).

1, 2, 2, 3, 5, 4, 7, 4, 5, 12, 4, 6, 14, 7, 6, 7, 21, 7, 6, 8, 23, 11, 14, 9, 26, 20, 14, 10, 28, 25, 24, 11, 39, 25, 24, 12, 41, 28, 28, 6, 12, 13, 54, 28, 28, 6, 12, 14, 56, 35, 42, 6, 12, 15, 59, 40, 57, 6, 12, 16, 61, 44, 65, 22, 12, 17, 78, 44, 65, 22, 12, 18, 80, 47, 71, 31, 30, 19, 99, 47, 71, 31, 30

Offset

1,2

Comments

Also, looking at all the partitions into equal-sized parts of all positive integers <= n, T(n,k) is the total number of parts in the partitions with the k-th largest parts.

Column k lists the partial sums of the column k of A027750.

The rows where the length row increases to a record gives A002182.

Links

Paolo Xausa, Table of n, a(n) for n = 1..10175 (rows 1..550 of triangle, flattened)

Michael De Vlieger, Log log scatterplot of row n, n = 1..5040, showing T(n, k) with trajectory of k in a color function where black indicates k = 1, red k = 2, ..., magenta k = 60, ignoring zeros.

Example

Triangle begins:
   1;
   2,   2;
   3,   5;
   4,   7,  4;
   5,  12,  4;
   6,  14,  7,   6;
   7,  21,  7,   6;
   8,  23, 11,  14;
   9,  26, 20,  14;
  10,  28, 25,  24;
  11,  39, 25,  24;
  12,  41, 28,  28,  6, 12;
  ...
For n = 6 the divisors, in increasing order, of all positive integers <= 6 are as follows:
  -----------------------------
  n\k |    1     2     3     4
  -----------------------------
  1   |    1
  2   |    1,    2
  3   |    1,    3
  4   |    1,    2,    4
  5   |    1,    5
  6   |    1,    2,    3,    6
.
The sum of the first divisors (k = 1) is equal to 1+1+1+1+1+1 = 6, so T(6,1) = 6.
The sum of the second divisors (k = 2) is equal to 2+3+2+5+2 = 14, so T(6,2) = 14.
The sum of the third divisors (k = 3) is equal to 4+3 = 7, so T(6,3) = 7.
The sum of the fourth divisors (k = 4) is equal to 6, so T(6,4) = 6.
So the 6th row of the triangle is [6, 14, 7, 6].
Also, for n = 6 the partitions into equal parts, with the sizes of the parts in decreasing order, of all positive integers <= 6 are as follows:
  ----------------------------------------------------
  n\k |     1      2              3           4
  ----------------------------------------------------
  1   |    [1]
  2   |    [2],  [1,1]
  3   |    [3],  [1,1,1]
  4   |    [4],  [2,2],       [1,1,1,1]
  5   |    [5],  [1,1,1,1,1]
  6   |    [6],  [3,3],       [2,2,2],   [1,1,1,1,1,1]
.
The total number of parts in the 1st partitions (k = 1) is 6, so T(6,1) = 6.
The total number of parts in the 2nd partitions (k = 2) is 14, so T(6,2) = 14.
The total number of parts in the 3rd partitions (k = 3) is 7, so T(6,3) = 7.
The total number of parts in the 4th partitions (k = 4) is 6, so T(6,4) = 6.
So the 6th row of the triangle is [6, 14, 7, 6].

Mathematica

nn = 20; s[_] = 0; m[0] = 0; Do[Set[m[n], Max[m[n - 1], DivisorSigma[0, n]]], {n, nn}]; Do[MapIndexed[(s[First[#2]] += #1; Set[t[n, First[#2]], s[First[#2]]]) &, PadRight[Divisors[n], m[n]]], {n, nn}]; Table[t[n, k], {n, nn}, {k, m[n]}] // Flatten (* Michael De Vlieger, Mar 04 2023 *)
A360461[rowmax_]:=DeleteCases[Accumulate[PadRight[Divisors[Range[rowmax]]]], 0, {2}];
A360461[20] (* Generates 20 rows *) (* Paolo Xausa, Mar 05 2023 *)

Prog

(PARI) rowlen(n) = vecmax(vector(n, k, numdiv(k))); \\ A070319
row(n) = my(vd=vector(n, i, divisors(i)), nb=rowlen(n)); vector(nb, k, sum(i=1, #vd, if (#(vd[i]) >= k, vd[i][k]))); \\ Michel Marcus, Mar 06 2023

Crossrefs

Row sums give A024916.

Row lengths give A070319.

Column 1 gives A000027.

Column 2 gives A088821.

The sum of the first n rows gives A175254.

Main sequences: A027750 and A244051.

Cf. A000005, A002182.

Sequence in context: A117267 A086363 A174094 * A284114 A323480 A330404

Adjacent sequences: A360458 A360459 A360460 * A360462 A360463 A360464

Keyword

nonn,tabf,look

Author

Omar E. Pol, Feb 07 2023

Status

approved